If $a$, $b$, and $c$ are sides of a triangle, then $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<2$. 
Let $a,b,c$ be the lengths of the sides of a triangle. Prove that
$$\sum_{\text{cyc}}\frac{a}{b+c}=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<2\,.$$

Attempt.  By clearing the denominators, the required inequality is equivalent to
$$a^2(b+c)+b^2(c+a)+c^2(a+b)>a^3+b^3+c^3\,.$$
Since $b+c>a$, $c+a>b$, and $a+b>c$, the inequality above is true.  Is there a better, non-bruteforce way?
 A: \begin{align*}
\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} & = \frac{2a}{2(b+c)}+\frac{2b}{2(c+a)}+\frac{2c}{2(a+b)} \\
&< \frac{2a}{a+b+c} + \frac{2b}{c+a+b} + \frac{2c}{a+b+c} \\
&= 2
\end{align*}
A: Since $a,b,c$ are the sides of a triangle, there exist $p,q,r\in\mathbb R^+$ such that $a=p+q$, $b=q+r$, $c=r+p$ (Ravi substitution).
$$\sum_{\text{cyc}}\frac{a}{b+c}<2\iff \sum_{\text{cyc}}\frac{p+q}{p+q+2r}<2$$
Now multiply both sides by $(2p+q+r)(p+2q+r)(p+q+2r)$, expand, rearrange.
See WolframAlpha (link) if you want.
$$\iff 3 p^2 q+3 p^2 r+3 p q^2+14 p q r+3 p r^2+3 q^2 r+3 q r^2
>0,$$
which is trivial, because $p,q,r>0$.
A: As @user236182 pointed out, we can write $a=p+q$, $b=q+r$, and $c=r+p$, with $p,q,r>0$. Letting $s = p+q+r$, we have
$$\frac{a}{b+c} = \frac{s-r}{s+r} = 1 - \frac{2r}{s+r} $$
and hence
$$\sum\limits_{\text{cyc}}{\frac{a}{b+c}}< 2 \iff \sum\limits_{\text{cyc}}{\left(1-\frac{2r}{s+r}\right)}< 2 \iff \sum\limits_{\text{cyc}}{\frac{2r}{s+r}} > 1. $$
By Cauchy-Schwarz we have
$$\left(\sum\limits_{\text{cyc}}{\frac{r}{s+r}}\right)\left(\sum\limits_{\text{cyc}}{r(s+r)}\right)\ge\left(\sum\limits_{\text{cyc}}{r}\right)^2 = s^2. $$
Furthermore,
$$\sum\limits_{\text{cyc}}{r(s+r)} = s\sum\limits_{\text{cyc}}{r}+\sum\limits_{\text{cyc}}{r^2} = s^2 + (p^2+q^2+r^2) < s^2+(p+q+r)^2 = 2s^2. $$
Note that the inequality is strict as $p,q,r>0$. It follows that
$$ 2s^2\sum\limits_{\text{cyc}}{\frac{r}{s+r}} > \left(\sum\limits_{\text{cyc}}{\frac{r}{s+r}}\right)\left(\sum\limits_{\text{cyc}}{r(s+r)}\right) \ge s^2 $$
and hence $\sum\limits_{\text{cyc}}{\frac{2r}{s+r}} > 1$, as desired.
A: Hint:  Without loss of generality, suppose $c$ is the largest side.  Hence, $c<a+b$.  Also,
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\leq\frac{a}{a+b}+\frac{b}{a+b}+\frac{c}{a+b}\,.$$
Note that the bound is sharp.  In the limit $a\to 0$ and $b\to c$, we have the sum goes to $2$.
A: 
Let $a,b,c$  be the lengths of the sides of a triangle. Prove that
\begin{align}\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}&<2\tag{1}\label{1}.\end{align}

Let $\rho,\ r,\ R$
be the semiperimeter,
inradius and circumradius
of the triangle $ABC$, then
we can rewrite \eqref{1} as
\begin{align} 
\frac a{2\rho-a}
+\frac b{2\rho-b}
+\frac c{2\rho-c}
&=
\frac{
4\,(a+b+c)\rho^2
-4\,(ab+bc+ca)\,\rho
+3\,abc
}{
8\rho^3-4\,(a+b+c)\,\rho^2+2\,(ab+bc+ca)\,\rho
-abc
}
<2
\tag{2}\label{2}
,
\end{align}
then, rearranging,
\begin{align} 
16\rho^3
-12(a+b+c)\rho^2
+8(ab+bc+ca)\rho-5abc
&>0
\tag{3}\label{3}
,\\
16\rho^3
-12(2\rho)\rho^2
+8(\rho^2+r^2+4rR)\rho-5\cdot4\rho r R
&>0
\tag{4}\label{4}
,\\
4r\rho(2r+3R)
&>0
\tag{5}\label{5}
,
\end{align}
which always holds.
